The classification of smooth equivalence relations is accomplished by two exercises from Su Gao’s book, Invariant Descriptive Set Theory.

Exercise 5.4.1, Su Gao, p 132.
Show that for any equivalence relation on a Polish space , iff there are infinitely many -equivalence classes.

Suppose . So there exists Borel such that iff . Since for each , the element lies in a different class, there are infinitely many classes.

Suppose there are infinitely many -classes; enumerate a countable subset: . By the Axiom of Choice, for each pick a , and define by . Since has the discrete topology, is cts, and so $latex \text{id}(\omega) \le_B E$.

Exercise 5.4.2, Su Gao, p 132
Show that if an equivalence relation is smooth then either , or , or for some finite , .

By Theorem 5.4.2, if is smooth, either (1) or (2) .

So assume (2). Then either has infinitely many classes or it does not.
If it does, by Exercise 5.4.1, , hence .

If it does not, suppose has classes. List the classes: . Since is smooth, each is Borel.

Then define by iff . Then is Borel. Also , since has discrete topology and we can pick
a point in each equivalence class to map to .

It’s been a while since my last post — I apologize to any devoted fans who have been disappointed.

I’m taking a class on Descriptive Set Theory now. Here is the proof of an interesting theorem from class: Assuming the Axiom of Choice, there exists an undetermined game.

Here, a game consists of two players, a pruned tree, and a payoff set . The players move alternately by picking an immediate extension of the last move. Player I wins if, after infinitely many moves, they have created an element of . Player II wins otherwise.

Kechris, p. 32, Exercise (6.6) (A)
Let be a Polish space. Decompose uniquely into , where is perfect, is countable open. This decomposition is possible by Cantor-Bendixson Theorem.

Exercise (6.6): is the largest perfect subset of .

Proof.
Suppose is perfect (so closed and perfect in its relative topology). From the proof of Theorem (6.4), and in the notation defined there, we have , hence . Now suppose , and further that ; notice that , being a set, is also a Polish space.

Claim: is a perfect space.
Let . Let be an open nbhd of . That is, there exists open in such that . Then is also open in , since is open in . Since is perfect, there exists . Therefore, is perfect and nonempty, hence uncountable by Theorem (6.2).

Kechris, p. 24
(B)
Let be a compact metrizable space and a metrizable space. We denote by the space of continuous functions from into with the topology induced by the uniform metric: where is a compatible metric for . A simple compactness argument shows that this topology is independent of the choice of .

Proof.
Let be two compatible metrics for . I will show that the two corresponding uniform metrics, give the same topology.

Suppose that did not give the same topology. That is, without loss of generality, there exists , , such that for all , . (Here .) That is ; that is, , which is the same as

and there exists such that.

Now, let . Let . Then there exists such that and . Since is compact, the sequence must have a convergent subsequence . So .

Let . Choose such that implies and . Then implies . So .

Therefore, converges to under . Thus, it must also do so under . But for all , . So this is a contradiction.